The parametric equation at any point $\theta$ of a curve is:
$$x=a\cos{(\theta)}+a\theta\sin{(\theta)}, y=a\sin(\theta)-a\theta\cos{(\theta)}$$
What is its distance from the origin to its normal at any $\theta$ ?
I don't understand how I should approach this problem and what steps should I follow. I got the equation to the curve ie $x^2+y^2=a^2{(1+θ^2)}$. but I don't understand what my next step should be. Please help.
Best Answer
The curve is an involute of circle with radius $a$.
Tangent of the curve: \begin{align*} \frac{dx}{d\theta} &= -a\sin \theta+a\theta \cos \theta+a\sin \theta \\ &= a\theta \cos \theta \\ \frac{dy}{d\theta} &= a\cos \theta+a\theta \sin \theta-a\cos \theta \\ &= a\theta \sin \theta \\[5pt] \frac{dy}{dx} &= \frac{a\theta \sin \theta}{a\theta \cos \theta} \\[5pt] &= \tan \theta \\ \end{align*}
Equation of the Normal: \begin{align*} y-(a\sin \theta-a\theta \cos \theta) &= -\cot \theta [x-(a\sin \theta+a\theta \cos \theta)] \\ y\sin \theta -a\sin \theta(\sin \theta-\theta \cos \theta) &= -x\cos \theta +a\cos \theta(\cos \theta+\theta \sin \theta) \\ x\cos \theta+y\sin \theta &= a (\sin^2 \theta+\cos^2 \theta) \\ x\cos \theta+y\sin \theta &= a \end{align*} which has fixed distance $a$ from the origin.